//

//解法1：dp[i]表示以i为结束位置，到i所需要花费的最小费用
class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();

        vector<int> dp(n + 1);
        dp[0] = 0, dp[1] = 0;

        for (int i = 2; i <= n; i++)
        {
            dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }

        return dp[n];
    }
};

//解法2：dp[i]表示以i为起始位置，到顶部所需要花费的最小费用
class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();

        vector<int> dp(n);
        dp[n - 1] = cost[n - 1], dp[n - 2] = cost[n - 2];

        for (int i = n - 3; i >= 0; i--)
        {
            dp[i] = min(dp[i + 1] + cost[i], dp[i + 2] + cost[i]);
        }

        return min(dp[0], dp[1]);
    }
};
